In: Advanced Math
A maximal plane graph is a plane graph G = (V, E) with n ≥ 3 vertices such that if we join any two non-adjacent vertices in G, we obtain a non-plane graph.
a) Draw a maximal plane graphs on six vertices.
b) Show that a maximal plane graph on n points has 3n − 6 edges and 2n − 4 faces.
c) A triangulation of an n-gon is a plane graph whose infinite face boundary is a convex n-gon and all of whose other faces are triangles. How many edges does a triangulation of an n-gon have?
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(b): Let G be a planar graph with n vertices and F faces and E number of edges .
Now , The boundary of every faces is a triangle , and each edge is on the boundary of two faces.
Therefore , if the number of edges on the boundary of a face is summed over all faces , the result will be 3F .
On the other hand, such a sum considers each edge twice .
So , 3F = 2E , i.e. , F=(2E)/3 and E=(3F)/2. ............... (i)
Applying Euler's formula ,
i.e., If a connected plane graph G has V vertices , E edges and F faces , then ,
. V-E+F = 2.
We obtain the desired results;
V - E + F = 2.
=> n - E + F = 2. ................ (ii)
Now , from (i) we have ,
n - E + (2E/3) = 2
=> 3n - 3E + 2E = 6
=> 3n - E = 6
=> E=3n-6
Similarly ,
From (i) and (ii) , we have ,
n - E + F = 2
=> n - (3F/2) + F = 2
=> 2n - 3F + 2F = 4
=> 2n - F = 4
=> F=2n-4.
(c) : Let R (n) be a graph whose vertices are every distinct triangulation of the n-gon , and whose edges represent the ability to get from one triangulation to another via diagonal flip. Let Nn be the number of vertices of R (n) .
Since , Number of edges of an n-gon is (n-3) .
Therefore , there are (n-3) flips that can be performed on each triangulation, which means that the degree of each vertex of R(n) =(n-3) , which means that R(n) is an (n-3)-gon graph.
Thus , Number of edges of R(n) is {Nn × (n-3)}/2. ( 2 in denominator is because each edge was considered twice).
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